Question

If ${\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(y\right)=\frac{2\pi }{3}$, then ${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(y\right)$ is equal to:

Answer 1

Computing the required value:

Given : ${\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(y\right)=\frac{2\pi }{3}$

$\begin{array}{l}\Rightarrow {\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(y\right)=\frac{2\pi }{3}\left[{\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)=\frac{\mathrm{\pi }}{2}\right]\\ \Rightarrow \frac{\pi }{2}–{\cos }^{-1}\left(x\right)+\frac{\pi }{2}-{\cos }^{-1}\left(y\right)=\frac{2\pi }{3}\\ \Rightarrow \pi –\frac{2\pi }{3}=co{s}^{-1}x+co{s}^{-1}y\\ \Rightarrow {\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(y\right)=\frac{\pi }{3}\end{array}$

Hence, the required value is $\frac{\pi }{3}$.